Conjugate Complex Zeros

    One of the nice properties of complex zeros of a polynomial with real coefficients is that if you know one complex zero (that is not a real number), you automatically know another! This follows from the theorem:
    Theorem Suppose f(x) is a polynomial with real coefficients and c = a + bi is a complex root of f(x). Then the conjugate of c, a - bi, is also a root of the polynomial.
Consequences:
  1. There must always be an even number of complex zeros.
  2. A polynomial with odd degree must have at least one real zero.
Example Consider the polynomial f(x) = x2 + 1. Clearly, this has no real roots, since x2 = -1 has no real solutions. However, the imaginary number i is a solution, hence also -i is a root of f(x).

Example Find all the roots of the polynomial  f(x) = x3 + x2 - 4x + 6.

First, we notice that the degree of the polynomial is 3, and odd number. Hence there will be at least one real zero for this polynomial. How can we find it? We can use, for example, the information from  Bounds on Real Roots of a Polynomial or  Rational Zeros of a Polynomial. The possible rational roots here are +1, +2, +3, and +6, and after checking them out we find that 2 is a root of this polynomial.

Now that we have one root of the polynomial, we divide f(x) by the linear factor that comes from this root: x - 2 (see  Reducing the Degree of a Polynomial), and find that

f(x) = (x - 2) (x2 - 2x + 2)
Thus in order to find the rest of the roots of  f(x), we need only find the roots of x2 - 2x + 2.

Using the Quadratic Formula we now find that x2 - 2x + 2 has the two conjugate complex roots

1 + i and 1 - i.

    Note that once we have a quadratic factor for a polynomial, using the quadratic formula will always give us both complex roots, since the conjugates are "built in" to the formula.